Divide the following complex numbers: $\dfrac{2(\cos(\frac{1}{6}\pi) + i \sin(\frac{1}{6}\pi))}{\cos(\frac{4}{3}\pi) + i \sin(\frac{4}{3}\pi)}$ (The dividend is plotted in blue and the divisor in plotted in green. Your current answer will be plotted orange.)
Dividing complex numbers in polar forms can be done by dividing the radii and subtracting the angles. The first number ( $2(\cos(\frac{1}{6}\pi) + i \sin(\frac{1}{6}\pi))$ ) has angle $\frac{1}{6}\pi$ and radius 2. The second number ( $\cos(\frac{4}{3}\pi) + i \sin(\frac{4}{3}\pi)$ ) has angle $\frac{4}{3}\pi$ and radius 1. The radius of the result will be $\frac{2}{1}$ , which is 2. The difference of the angles is $\frac{1}{6}\pi - \frac{4}{3}\pi = -\frac{7}{6}\pi$ The angle $-\frac{7}{6}\pi$ is negative. A complex number goes a full circle if its angle is increased by $2 \pi$ , so it goes back to itself. Because of that, angles of complex numbers are convenient to keep between $0$ and $2 \pi$ $-\frac{7}{6}\pi + 2 \pi = \frac{5}{6}\pi$ The radius of the result is $2$ and the angle of the result is $\frac{5}{6}\pi$.